1. Section 1.2
Trigonometric Ratios

2. Objectives:
1. To state and apply the Pythagorean theorem.
2. To define the six trigonometric ratios.

3. Pythagorean Theorem
In right ABC, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse.
a2 + b2 = c2 B A C

4. Trigonometric Ratios leg opposite A sine of A = hypotenuse
leg adjacent A
cosine of A =
leg adjacent A
leg opposite A
tangent of A =

5. SOHCAHTOA ine pposite ypotenuse osine djacent ypotenuse angent pposite

6. Trigonometric Ratios h o A sin = h a A cos = a o A tan = hypotenuse
opposite
adjacent A h o A
sin = h a A
cos = a o A
tan =

7. Reciprocal Ratios 1 cosecant of A = cscA = sinA 1
cosA 1
secant of A = secA =
tanA 1
cotangent of A = cotA =

8. Reciprocal Ratios o h A csc = a h A sec = o a A cot = A opposite
adjacent
hypotenuse o h A
csc = a h A
sec = o a A
cot =

9. EXAMPLE 1 Find the six trigonometric ratios for G in right EFG.
g2 + e2 = f2
g = 82
g = 64
g2 = 28
g =
2 7 8 6 E F

10. Practice Question: Find the six trigonometric ratios for E in right EFG.9 11 G E F
sin E =
1. 2.
3. 4. 9 11
2 10 11
9 10 20
2 10 9

11. P(x,y) r y  x

12. Trigonometric Ratios r y hyp opp sin = q y r opp hyp csc = q r x hyp
adj
cos = q x r
adj
hyp
sec = q x y
adj
opp
tan = q y x
opp
adj
cot = q

13. EXAMPLE 2 Find the six trigonometric ratios for a 90º angle.
x=0, y=1, r=1
cos  = 0
sin  = 1
tan  = und.
90° P 

14. EXAMPLE 2 Find the six trigonometric ratios for a 90º angle.
x=0, y=1, r=1
sec  = und.
csc  = 1
cot  = 0
90° P 

15. Practice Question: Find the six trigonometric ratios for a 180º angle.
sin  =
und.
180°  P

16. Special Triangles a2 + b2 = c2 12 + 12 = c2 1 + 1 = c2 c2 = 2 45 1 2 c

17. Special Triangles 2 ÷ ø ö ç è æ 2 1
sin 45° = 2 c = 45 1

18. Special Triangles 2
sin 45° = 2 c = 45 1 2
cos 45° =
tan 45° = 1

19. Special Triangles
csc 45° = 2 2 c = 45 1
sec 45° = 2
cot 45° = 1

20. 2
sin 45° =
cos 45° =
tan 45° = 1
csc 45° =
sec 45° =
cot 45° =

21. Special Triangles a2 + b2 = c2 12 + b2 = 22 1 + b2 = 4 b2 = 3 60 2 130 3 b =

22. Special Triangles 3 ÷ ø ö ç è æ 3 1 60 2 1 30
tan 30° = 3 b =

23. Special Triangles 60 2 1 30 3 b = tan 30° = 3 1 sin 30° = 2 3
cos 30° =

24. Special Triangles 60 2 1 30 3 b = cot 30° = 3 csc 30° = 2 3 2
sec 30° =

25. 3
tan 30° =
sin 30° =
cos 30° = 2 1
cot 30° =
csc 30° =
sec 30° =

26. Special Triangles 60 2 1 30 3 b = 2 3 sin 60° = 2 1 cos 60° =
tan 60° = 3

27. Special Triangles 60 2 1 30 3 b = 3 2 csc 60° = sec 60° = 2 3
cot 60° =

28. sin 60° =
cos 60° =
tan 60° = 2 3 1
csc 60° =
sec 60° =
cot 60° =

29. Homework: pp

30. ►A. Exercises
1. Find the six trig. ratios for both acute angles in each triangle. A 5 2 B C 21

31. 1st-solve for side n n2 = 122-22 = 144-4 = 140 22 + n2 = 122 35 2 140
►A. Exercises
3. Find the six trig. ratios for both acute angles in each triangle. n M L 2 12 N
1st-solve for side n
n2 = = = 140
22 + n2 = 122 35 2
140 n =

32. #3. 35 2 M L 2 12 N

33. ►A. Exercises
7. Find the six trig. functions for an angle in standard pos. whose terminal ray passes through the point (-6, -1). -6  -1 37

34. -6 -1 37 
sin  = csc  =
cos  = sec  =
tan  = cot  =

35. ►B. Exercises
15. Find the six trig. ratios for the quadrantal angle measuring 180°.

36. ►B. Exercises
15. Find the six trig. ratios for the quadrantal angle measuring 180°.
(-1, 0)

37. ■ Cumulative Review
30. Give the distance between (2, 7) and
(-3, -1).

38. ■ Cumulative Review
31. Give the midpoint of the segment joining (2, 7) and (-3, -1).

39. ■ Cumulative Review
32. Give the angle  coterminal with 835 if 0    360.

40. ■ Cumulative Review
33. Convert 88 to radians.

41. ■ Cumulative Review
34. If sec  = 7, find cos .